Question

Verify that force fields is conservative. Then find, a scalar potential $\mathit{\phi }$such that $\mathit{F}\mathbf{=}\mathbf{-}\mathbf{\nabla }\mathit{\phi }\mathbf{,}\mathbf{}\mathit{F}\mathbf{}\mathbf{=}\mathbf{}\mathit{i}\mathbf{}\mathbf{-}\mathbf{}\mathit{z}\mathit{j}\mathbf{}\mathbf{-}\mathbf{}\mathit{y}\mathit{k}$ .

Short answer

The force field is conservative and the scalar potential $\phi =-x+yz$

Step-by-step solution

Given Information

The given force vector is

$F=i-zj-yk$.

Definition

Force field is said to be conservative if curl of force becomes zero.

Concept and Formula

The formula to find $\nabla \times F$ in order to show that field is conservative is mentioned below.

$\nabla \times F=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x& y& z\end{array}\right|$

where $F=xi+yj+zk$

In order to find the scalar potential $\phi$,where $F=-\nabla \phi F=-\nabla \phi$, the formula for work done is given below.

$$\begin{gathered} W=\int F.dr \\ F=\nabla W \\ \Rightarrow \phi =-W \end{gathered}$$

Conservative force field

Use the formula $\nabla \times F=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x& y& z\end{array}\right|$

Put $F=i-zj-yk$

$$\begin{gathered} \nabla \times F=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 1& -z& -y\end{array}\right| \\ \nabla \times F=\left(-1+1\right)i-\left(0-0\right)j+\left(0-0\right)k \\ \nabla \times F=0 \end{gathered}$$

This implies that force field is conservative.

Calculate Work done

Use the formula $W=\int F.dr$

$W=\int dx-zdy-ydz$

Take path from (0,0,0) to (x,y,z)

(i) From (0,0,0) to (x,0,0)

y = 0

z = 0

dy = 0

dz = 0

$$\begin{gathered} W_1=\underset{0}{\overset{x}{\int }}dx \\ {W}_1=x \end{gathered}$$

(ii) From (x,0,0) to (x,0,z)

y = 0

dy = 0

dx = 0

$$\begin{gathered} W_2=\underset{0}{\overset{z}{\int }}0-z\left(0\right)-0\left(dz\right) \\ {W}_2=0 \end{gathered}$$

(iii) From (x,0,z) to (x,y,z)

dx = 0

dz = 0

$$\begin{gathered} W_3=\underset{0}{\overset{y}{\int }}0-zdy-y\left(0\right) \\ {W}_3=-zy \end{gathered}$$

Total work done is given below.

$$\begin{gathered} W=W_1+W_2+W_3 \\ W=x+0-zy \\ W=x-zy \end{gathered}$$

Calculate scalar potential

Use the formula $F=\nabla W$

Also, use the given relation between force and scalar potential which is given below.

$F=-\nabla \phi$

From both the above formulas we get that $\phi =-W$

$$\begin{gathered} \Rightarrow \phi =-\left(x-yz\right) \\ \Rightarrow \phi =-x+yz \end{gathered}$$

The force field is conservative and the scalar potential $\phi =-x+yz$